Why the Numerator Is a Targeting Choice
Inverse probability of censoring weights compensate for the selective loss of follow-up: histories that are unlikely to remain observed receive more weight. Stabilizing the weights adds a numerator, often based on baseline covariates, to reduce variability. That numerator is not merely a harmless scaling factor. Across follow-up, it defines a reference censoring process, and therefore affects the distribution represented by weighted records.
The key distinction is between preserving a baseline distribution and preserving a conditional distribution. A constant numerator leaves the normalized baseline distribution unchanged under the ideal weighting identity. A numerator that varies with baseline covariates generally tilts their marginal distribution among weighted records. It can, however, leave the distribution of other baseline variables conditional on those numerator covariates unchanged.
The denominator conditions on the history needed to address informative censoring. The numerator uses less information, here only the baseline vector \(V\), which may be empty or a subset of \(Z\). For a probability interpretation, take \(q_k(V)\) to be valid retention probabilities, with \(0<q_k(V)\leq 1\) on the population of interest. The product \(Q_t(V)=\prod_{k=1}^{t}q_k(V)\) is the numerator's probability of remaining uncensored through \(t\) under its reduced conditioning set.
The distributional identities below are population statements. They assume consistency of the observed follow-up process and sequential independent censoring: at each time, conditional on the observed history of those still at risk, the censoring decision is independent of their future complete-data history. They also require censoring positivity, \(g_k(H_{k-1})>0\) on histories with positive probability, and finite follow-up \(t\). The numerator probabilities must be positive wherever a weighted target is intended. These conditions allow the denominator to remove selection due to censoring; they do not imply that the numerator leaves every baseline distribution unchanged.
The Distribution Induced by the Numerator
Under these assumptions, the denominator cancels the actual retention process. What remains is the numerator process. For any integrable function \(h(Z)\), the weighted expectation among records observed through \(t\) is
In particular, the total expected weight is \(E[W_t]=E[Q_t(V)]\). Thus, after normalizing the weighted records to have total mass one, their baseline distribution is the distribution tilted by \(Q_t(V)\). This result is about normalized weighted records; it does not say that the unnormalized total weight equals one.
To prove the theorem, condition on the complete-data history and apply the censoring assumptions sequentially. At each time \(k\), among individuals still at risk, the conditional expectation of the retention indicator divided by its true retention probability is
Sequential independent censoring makes this cancellation valid when the histories are successively integrated over the complete-data process. Multiplying the factors over \(k=1,\ldots,t\) cancels the true retention process. The numerator product is a function of baseline \(V\), so it remains in the expectation. Consequently, \(E[W_t h(Z)]=E[Q_t(V)h(Z)]\) for integrable \(h\). Taking \(h(Z)=I(Z\in A)\), and then \(h(Z)=1\), yields the stated ratio. The discrete formula follows by choosing \(A=\{z\}\); when a density exists, integrating over arbitrary sets gives the density formula. \(\square\)
This result precisely describes what “preserved” means. If \(V\) is a subset of \(Z\) and \(Q_t(V)>0\), then for any measurable set \(B\),
Indeed, in the ratio defining the weighted conditional probability, \(Q_t(V)\) is constant at a fixed value of \(V\), so it cancels. But the marginal distribution of \(V\) itself is generally tilted:
The numerator therefore preserves the conditional composition of \(Z\) within levels of \(V\), not necessarily the original marginal distribution of \(V\). If \(Q_t(V)\) is constant, the tilt cancels from every baseline set and the full marginal distribution of \(Z\) is preserved.
Worked Example: A Baseline-Dependent Numerator Tilts a Binary Covariate
Suppose \(Z\) is binary, \(P(Z=1)=0.60\), and there is one follow-up interval. Let the numerator retention probabilities be \(q_1(0)=0.80\) and \(q_1(1)=0.50\). Then \(Q_1(Z)=q_1(Z)\), and $$ E[Q_1(Z)]=(0.40)(0.80)+(0.60)(0.50)=0.32+0.30=0.62. $$
The weighted proportion with \(Z=1\) is $$ P_W(Z=1)=\frac{(0.60)(0.50)}{0.62}=\frac{0.30}{0.62}\approx 0.484. $$ The original proportion was \(0.60\), so this numerator does not preserve the marginal distribution of \(Z\). The weighted proportion with \(Z=0\) is \(0.32/0.62\approx 0.516\); the two probabilities sum to one. The tilt is caused by the numerator's lower retention probability for \(Z=1\), not by a failure of the denominator to correct informative censoring.
Worked Example: A Constant Numerator Preserves the Baseline Distribution
Let \(Z\) again be binary with \(P(Z=1)=0.60\), but set the one-period numerator retention probability to \(q_1(Z)=0.70\) for both values. Then \(E[Q_1(Z)]=0.70\), and $$ P_W(Z=1)=\frac{(0.60)(0.70)}{0.70}=0.60,\qquad P_W(Z=0)=\frac{(0.40)(0.70)}{0.70}=0.40. $$
The overall expected weight is \(0.70\), not one, because \(E[W_1]=E[Q_1(Z)]\). Yet after normalizing by that expected weight, the baseline distribution is exactly the original distribution. This example separates two facts that are sometimes confused: constant numerators need not make the weights have mean one, but they do preserve the normalized baseline distribution under the theorem's assumptions.
Worked Example: The Density Tilt for a Continuous Baseline Variable
Let \(Z\) be uniform on \([0,1]\), so \(f_Z(z)=1\) on that interval. Choose a one-period numerator \(q_1(z)=0.4+0.4z\), which is a valid retention probability between \(0.4\) and \(0.8\). Its mean is $$ E[Q_1(Z)]=\int_0^1(0.4+0.4z)\,dz=0.4+0.2=0.6. $$
The weighted density is therefore $$ f_{W,Z}(z)=\frac{1(0.4+0.4z)}{0.6}=\frac{2}{3}(1+z),\qquad 0\leq z\leq 1. $$ It integrates to one because \(\int_0^1 \frac{2}{3}(1+z)\,dz=\frac{2}{3}(1+\frac12)=1\). Its mean is $$ E_W[Z]=\int_0^1 z\frac{2}{3}(1+z)\,dz =\frac{2}{3}\left(\frac12+\frac13\right)=\frac{5}{9}\approx 0.556. $$ The original mean is \(1/2\), so the numerator tilts the continuous marginal distribution toward larger values of \(Z\). A formula based on point probabilities would not express this result, since \(P(Z=z)=0\); the density or set-based formula is required.
How the Choice Changes the Weighted Target
A constant numerator and a baseline-dependent numerator answer different targeting questions. With a constant numerator, normalized weighted records represent the original baseline population, assuming the censoring correction is valid. With \(V\)-dependent numerators, they represent a baseline population reweighted in proportion to \(Q_t(V)\). This can be useful when the analysis is intended to retain a selected baseline structure, but the resulting marginal summaries should not automatically be described as summaries for the original population.
For an outcome \(Y_t\) observed only while uncensored, the same sequential weighting argument gives \(E[W_tY_t]=E[Q_t(V)Y_t]\), under the same censoring assumptions and with \(Y_t\) understood as its complete-data value. Hence the normalized weighted mean targets
If \(Q_t(V)\) is constant, this is \(E[Y_t]\). If it varies with \(V\), it is a tilted mean, generally different from the original marginal mean. A conditional outcome model can describe outcomes at fixed \(V\), but a marginal prediction still requires an explicit choice of the distribution over which to average. The chosen numerator alone does not make a fitted marginal effect equal to the effect standardized to the original baseline population.
A baseline-dependent numerator can also be motivated by a desire for a less variable weight distribution than one obtained from a constant numerator. That is a variance consideration, not a guarantee of the original-population target. Choosing a numerator therefore requires stating both the covariates it conditions on and the population distribution intended for the final interpretation.
Diagnostics, Alternatives, and Limitations
Diagnostics should check whether the implemented weights behave as the stated numerator implies. Compare the weighted distribution of \(V\) with the distribution tilted by \(Q_t(V)\), and compare the weighted distribution of other baseline variables conditional on \(V\) with their unweighted conditional distributions. For a constant numerator, normalized weighted baseline distributions should match the original ones in the population idealization. In finite samples, discrepancies can occur through sampling variability, estimated weights, or model error.
Also inspect the range and distribution of the weights, their mean relative to the average numerator survival probability, and an effective sample size such as \((\sum_i W_{it})^2/\sum_i W_{it}^2\). Very large weights or a small effective sample size indicate that a few observed records dominate. Such instability may reflect weak censoring positivity or poor estimation of the denominator; a convenient numerator cannot repair a zero or near-zero true retention probability. If the weighted baseline distribution fails to follow the intended tilt, check the numerator implementation, the time indexing, and whether the denominator and numerator products cover the same intervals.
| Numerator choice | Normalized baseline distribution | Interpretation |
|---|---|---|
| Constant \(q_k\), not depending on \(Z\) | Original marginal distribution of \(Z\) | Targets the original baseline population after normalization |
| Depends on \(V\subseteq Z\) | Marginal distribution of \(V\) is tilted by \(Q_t(V)\); \(Z\mid V\) is unchanged | Targets a population weighted across \(V\) strata |
| Arbitrary positive factor not interpretable as retention probabilities | Weighted distribution is tilted by its product factor, if the identity applies | May be usable as a weighting scheme, but lacks the stated numerator survival interpretation |
The first two rows are alternatives, not competing corrections for different kinds of censoring: both use the same denominator to address informative loss to follow-up. Their difference is the reference population induced by the numerator. A further limitation is that the theorem describes the target under correct denominator probabilities and the stated assumptions. A misspecified denominator need not cancel the actual censoring process, so the resulting weighted distribution may not equal the intended tilt. Nor does a favorable baseline balance diagnostic prove sequential independent censoring or positivity.
Common Mistakes
- Claiming every stabilized weight preserves the original baseline distribution. That conclusion holds for a numerator product constant in baseline covariates, after normalization. If it varies with \(V\), the marginal distribution of \(V\) is tilted.
- Confusing conditional preservation with marginal preservation. A numerator depending on \(V\) leaves \(Z\mid V\) unchanged under the identity, but that does not mean it leaves the marginal distribution of \(Z\) unchanged.
- Using a point-probability formula for a continuous covariate. For continuous \(Z\), use \(P_W(Z\in A)\) for measurable sets or the density formula when a density exists; point probabilities are zero.
- Treating the numerator as a correction for positivity failure. A numerator changes the target and can affect weight variability, but it cannot supply information for histories with zero probability of remaining uncensored.
- Reporting a weighted marginal mean without specifying its population. A baseline-dependent numerator can make that mean a tilted-population summary rather than the original-population mean.
Check Your Understanding
Answer each question using the distinction between a marginal distribution and a conditional distribution.
- If \(Q_t(V)\) is constant, what is the normalized weighted distribution of \(Z\)?
- Suppose \(V\) is a subset of \(Z\), and \(Q_t(V)>0\). Which distribution is preserved, and which may be tilted?
- For continuous \(Z\), what expression gives its weighted density when \(V=Z\) and a density exists?
- In the binary example with \(P(Z=1)=0.60\) and \(q_1(1)=0.50,\ q_1(0)=0.80\), calculate the weighted proportion with \(Z=1\).
- Why can a baseline-dependent numerator not fix a violation of censoring positivity?
Solutions. (1) The original marginal distribution of \(Z\), because the constant \(Q_t\) cancels from the normalized set probabilities. (2) \(Z\mid V\) is preserved; the marginal distribution of \(V\) may be tilted in proportion to \(Q_t(V)\). (3) \(f_{W,Z}(z)=f_Z(z)Q_t(z)/E[Q_t(Z)]\). (4) The numerator is \((0.60)(0.50)=0.30\), and the normalizing expectation is \((0.60)(0.50)+(0.40)(0.80)=0.62\), giving \(0.30/0.62\approx0.484\). (5) Positivity requires a positive true retention probability on relevant histories. If that probability is zero, the data contain no uncensored observations from that history for the denominator to reweight; changing the numerator cannot create them.